It might be a little easier to read if pull the constants (in the numerator) outside of the integral, and then apply a `U substitution`.\[\large \int\limits\limits_{0}^{t} \frac{Vm}{M-mt} dt \qquad \rightarrow \qquad Vm\int\limits\limits_{0}^{t} \frac{1}{M-mt}dt\]
Are you familiar with U substitution? :)

Yes, you can bring the negative into the log as an exponent. Which will give us this,\[\large a \ln\left[\left(\frac{c-bt}{c}\right)^{-1}\right] \qquad = \qquad a \ln\left[\left(\frac{c}{c-bt}\right)\right]\]
Whichever way makes sense to you though :3

ok, so f has to be in there cuz that's what i'm trying to solve. so plugging in the other letters and simplifying, according to the solution, it's suppose to be
a[ln(1+(f/e))] but i can't get it. however, i haven't tried with your second expression up there so i will try it now